Finding the length of a piece of string using Lagrange multipliers

Question

A string of fixed length L is to be cut into three pieces of length x, y and z. The string of length x is to be shaped into a circle, the string of length y is to be shaped into a square and the string of length z is to be shaped into triangle with angles π/2, π/4, π/4. How do I use Lagrange multipliers to find the values of x, y and z where local minima or maxima may occur.

Answer 1

to maximize area, it should be clear that one shape has a higher area / perimeter ratio than multiple shapes. Furthermore, the circle more efficient than the other shapes. So all of the string in one circle maximizes area.

to minimize area.

$2\pi r = x\\ r = \frac {x}{2\pi}$

Area of a circle $= \pi r^2 = \frac{x^2}{4\pi}$

sqrare:

$4s = y\\ s = \frac {y}{4}$

Area $= s^2 = \frac {y^2}{16}$

Triangle

$(2 + \sqrt 2)s = z\\ s = \frac {z}{2 + \sqrt 2}$

Area $= \frac 12 s^2 = \frac {z^2}{12+8\sqrt{2}}$

minimize: $\frac {1}{4\pi} x^2 + \frac {1}{16} y^2 + \frac 1{12+8\sqrt2} z^2$

constrained by $x+y+z = L$

$f(x,y,z,\lambda) = \frac {1}{4\pi} x^2 + \frac {1}{16} y^2 + \frac 1{12+8\sqrt2} z^2 - \lambda (x+y+z-L)$

$\frac {\partial f}{\partial x} = \frac {2}{4\pi} x - \lambda = 0\\ \frac {\partial f}{\partial y} = \frac {2}{16} y - \lambda = 0\\ \frac {\partial f}{\partial z} = \frac 2{12+8\sqrt2} y - \lambda = 0\\ \frac {\partial f}{\partial \lambda} = x+y+z - L = 0$

And this is a system of linear equations which you learned how to solve in basic algebra.